anonymous
  • anonymous
How do I use the exponential shift to solve the linear differential equation y'' + 7y' +12y = 2(e^-3)(cos2x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
im looking to find the general solution y = yc +yp. so far i have solved the homogenous part of the equation yc. yc = Ae^-3x + Be^-4x. i cant solve for yp yet.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Let me do it on paper first. I think it would be murder typing it out.

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anonymous
  • anonymous
haha thank you! any help will go a long way
anonymous
  • anonymous
i'll continue to try draw something out of my mess of numbers and letters while i wait
anonymous
  • anonymous
i also figured that the derivative operator 'D' helps so that the LHS = (D^2 + 7D +12)y
anonymous
  • anonymous
You do mean e^{-3x}, don't you?
anonymous
  • anonymous
yes! sorry typo
anonymous
  • anonymous
When I get to factors that need expanding, I'm going to get Mathematica to do that (save time).
anonymous
  • anonymous
ok no worries. if i dont follow il just let u know
anonymous
  • anonymous
So what happened, your lecturer just dump theory on you with no demonstration?
anonymous
  • anonymous
he demonstated with the d.e. y'' + 6y' + 13y = x(e^3x)(sin2x) however that takes y = imaginary(Y) and also takes into account the complex numbers. As my roots are reall -3, -4 and i take y=real(Y) instead, i dont know what yp to try
anonymous
  • anonymous
he pretty much chalk vomits all over the board without going into the specifics of how he gets to places
anonymous
  • anonymous
Not trained to teach...
anonymous
  • anonymous
anonymous
  • anonymous
its really quite frustrating... luckily we are changing lecturers next week.. sorry for any hassle im causing you!
anonymous
  • anonymous
just give me a fan point and i'll be happy
anonymous
  • anonymous
wow thanks! im just gona go through it and check i understand.. haha ok sure how do i do that?
anonymous
  • anonymous
just click my 'become a fan' link...if there isn't one there, refresh the page...should work then
anonymous
  • anonymous
ps. is there a place to download mathmatica for free? ok right on it
anonymous
  • anonymous
www.wolframalpha.com The guy who made mathematica made this site. It can do a lot of things like integration, solving differential equations, plotting, etc.
anonymous
  • anonymous
fanned. thanks a million! will i still be able to ask help from you if i have any more troubles..
anonymous
  • anonymous
this is wolfram's solution to your equation http://www.wolframalpha.com/input/?i=y%27%27-2y%27%2B5y+%3D+16x^3*e^%283x%29+++
anonymous
  • anonymous
woops...no it's not...wait.
anonymous
  • anonymous
haha! i was like crap.. i did notttttt get that
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=y%27%27%2B7y%27%2B12y+%3D+2e^%28-3%29cos%282x%29
anonymous
  • anonymous
haha ur awesome at pen to paper but im afraid you did the same typo as me last time. theres an 'x' in e^-3x. however the graphs dont look how i thought they would..
anonymous
  • anonymous
graphs in the correct version*
anonymous
  • anonymous
ah...typo...
anonymous
  • anonymous
yep :) i thought it would have more of a decaying oscillation as it is overdamped.. is it because of the negative exponential? why?
anonymous
  • anonymous
So, am I right in understanding you've finished deriving the solution?
anonymous
  • anonymous
And you're reading off that and not Wolfram?
anonymous
  • anonymous
sorry no. i was comparing my notes to the graphs of wolfram.. but! i havent read through the wolfram content yet.. just making observations. if you would like you can continue helping other students.. and im guessing when i need help again it will notify you?
anonymous
  • anonymous
Yes, I get an e-mail if you respond to a thread I've been in. Are you fine to derive the rest?
anonymous
  • anonymous
If an error's been made on my part, it's because it's after 1am where I am. If you need help in the next ~12-24hrs, I may not be around. You'd should try your luck with someone else.
anonymous
  • anonymous
i should be ok for now thanks :) it takes me quite a while longer than you to come up with a solution
anonymous
  • anonymous
yep me too. 1:22am. ok do u recommend anyone else?
anonymous
  • anonymous
I don't know the particular skill level of people on here. You may have to post and cross fingers.
anonymous
  • anonymous
Is this an assignment due tomorrow or something?
anonymous
  • anonymous
ahah yeh you caught me out. due 2pm. i wish i could say im a more dedicated student working til the early hours of the morning doing maths... ah the thrills of life!
anonymous
  • anonymous
Well, since your timezone's mine, if I see something from you in the morning and I can respond, I will. My day will be hell busy, though.
anonymous
  • anonymous
Anyways, good luck. I seriously have to hit the sack.
anonymous
  • anonymous
thanks you are a hero! night!

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